Answer:
Step-by-step explanation:
In the figure attached,
ΔABC is an equilateral triangle,
Sides AB = BC = AC and points P, Q, and R are the midpoints of these sides respectively.
If the coordinates of A(0, 0), B(2a, 0) and C(a, b)
AB = 2a
AC = [tex]\sqrt{a^2+b^2}[/tex]
Since AB = AC
2a = [tex]\sqrt{a^2+b^2}[/tex]
4a² = a² + b²
3a² = b²
Therefore, ordinate pairs representing midpoints of AB, BC and AC will be
P = [tex](\frac{2a+0}{2},\frac{0}{2})[/tex] =(a, 0)
Q = [tex](\frac{a+2a}{2},\frac{b}{2})[/tex] = [tex](\frac{3a}{2},\frac{b}{2})[/tex]
R = [tex](\frac{a+0}{2},\frac{b+0}{2})[/tex] = [tex](\frac{a}{2},\frac{b}{2})[/tex]
Now we will find the lengths of medians with the help of formula of distance between two points (x, y) and (x', y')
d = [tex]\sqrt{(x-x')^2+(y-y')^2}[/tex]
AQ = [tex]\sqrt{(0-\frac{b}{2})^2+(0-\frac{3a}{2})^2}[/tex]
= [tex]\sqrt{\frac{b^2}{4}+\frac{9a^2}{4}}[/tex]
= [tex]\frac{1}{2}(\sqrt{b^2+9a^2})[/tex]
= [tex]\frac{1}{2}\sqrt{12a^2}[/tex] [Since b² = 3a²]
= [tex]a\sqrt{3}[/tex]
BR = [tex]\sqrt{(2a-\frac{a}{2})^{2}+(0-\frac{b}{2})^2}[/tex]
= [tex]\sqrt{(\frac{3a}{2})^2+(-\frac{b}{2})^2}[/tex]
= [tex]\frac{1}{2}\sqrt{b^2+9a^2}[/tex]
= [tex]\frac{1}{2}(\sqrt{12a^2})[/tex]
= [tex]a\sqrt{3}[/tex]
CP = b = [tex]a\sqrt{3}[/tex]
Therefore, AQ = BR = CP = [tex]a\sqrt{3}[/tex]
Hence, medians of an equilateral triangle are equal.
